Int. J. Anal. Appl. (2023), 21:34

Solvability of Nonlinear Wave Equation with Nonlinear Integral Neumann Conditions

Iqbal M. Batiha1,2,∗, Zainouba Chebana3, Taki-Eddine Oussaeif3, Adel Ouannas3,

Iqbal H. Jebril1, Mutaz Shatnawi4

1Department of Mathematics, Al-Zaytoonah University of Jordan, Amman 11733, Jordan
2Nonlinear Dynamics Research Center (NDRC), Ajman University, Ajman 346, UAE

3Department of Mathematics and Computer Science, University of Larbi Ben M’hidi, Oum El

Bouaghi, Algeria
4Department of Mathematics, Irbid National University, Irbid 21110, Jordan

∗Corresponding author: i.batiha@zuj.edu.jo

Abstract. In this paper, we examine a nonlinear hyperbolic equation with a nonlinear integral condition.

In particular, we prove the existence and the uniqueness of the linear problem by the Fadeo Galerkin

method, and by applying an iterative process to some significant results obtained for the linear problem,

the existence and the uniqueness of the weak solution for the nonlinear problem are additionally

examined.

1. Introduction

The nonlinear hyperbolic equations describes important processes in the nonlinear evolution equation

basis of mathematical models of diverse phenomena and processes in mechanics, physics, technology,

biophysics, biology, ecology, and many other areas [1–4]. Such ubiquitous occurrence of nonlinear

hyperbolic equations is to be explained, first of all, by the fact that they are derived from fundamental

laws in the real world [5,6].

Let us remark only that for broad classes of equations, the fundamental questions of solvability and

uniqueness of solutions of various boundary value problems have been solved and that the differentia-

bility properties of the solutions have been studied in detail [7–9]. General results of the solvability

and uniqueness were inferred by different methods such as the energy method, upper lower method

Received: Dec. 25, 2022.

2010 Mathematics Subject Classification. 35D30, 35L05, 35L70.

Key words and phrases. hyperbolic equation; nonlinear integral condition; weak existence.

https://doi.org/10.28924/2291-8639-21-2023-34
ISSN: 2291-8639

© 2023 the author(s).

https://doi.org/10.28924/2291-8639-21-2023-34


2 Int. J. Anal. Appl. (2023), 21:34

and the Fadeo Galerkin methods. The later one is regarded one of the most important methods that

were mainly developed in the 1960s, but they are still powerful tools today to deal with nonlinear evo-

lution equations, especially those who are modeled by non-classical boundary conditions that consist

of integral conditions [10–13]. Non-local and integral partial differential equations are used to solve a

vast range of current physics and technology challenges [14–19]. When it is hard to directly measure

the minimum and maximum values on the border, the overall value or average is known. This method

might be utilized for modeling where we can model more complicated domain with nonlinear integral

condition.

Motivated by the above perspective, we trait in this work to discuss a nonlinear evolution equation

with a nonlinear integral condition. In particular, we aim to focus on the solvability of the solution

of nonlinear hyperbolic problems with the integral condition of the second type by the method of

Fadeo-Galerkin. In the following sections, we present first the existence of the linear problem, and

then by applying an iterative process based on the results obtained for the linear problem, we prove

the existence and the uniqueness of the weak solution of the nonlinear problem.

2. The statement of the main problem

In this section, we let Q =
{

(x,t) ∈R2, x ∈ Ω = ]0, l[ and 0 < t < T
}
, besides we consider the

main following initial boundary value problem for a nonlinear hyperbolic equation:


∂2u

∂t2
−a

∂2u

∂x2
= f (x,t,u,ux )

u(x, 0) = ϕ(x)

ut(x, 0) = ψ(x)
∂u

∂x
(0,t) =

∫ l
0
k(x,t)g(ut)(x,t)dx

∂u

∂x
(l, t) =

∫ l
0
k(x,t)h(ut)(x,t)dx.

(P1)

Assuming that f ∈ L2 (Q) and ϕ,ψ ∈ L2 (Ω). The nonlinear hyperbolic equation is given as follows:

Lu =
∂2u

∂t2
−a

∂2u

∂x2
= f (x,t,u,ux ), (2.1)

which satisfies the following identities:

• The initial conditions

`u =

{
u(x, 0) = ϕ (x)

ut (x, 0) = ψ (x)
, x ∈ (0, l) .

• The boundary conditions are integral conditions of the second type defined as:

∂u

∂x
(0,t) =

∫ l
0

k(x,t)g(ut)(x,t)dx, t ∈ (0,T ) ,

∂u

∂x
(l, t) =

∫ l
0

k(x,t)h(ut)(x,t)dx, t ∈ (0,T ) ,



Int. J. Anal. Appl. (2023), 21:34 3

where

k(x,t) ≥ 0 ∀(x,t) ∈ Qand g(ut)(x,t) ≤ h(ut)(x,t) ∀(x,t) ∈ Q,

and for all v ∈ L2(Q), we have:

‖g(v)‖L2(Q) ≤ C‖v‖L∞(0,T,L2(Ω)) ,

in which we define the space V by V = H1 (Ω).

Actually, the space V is provided with the norm ‖v‖V = ‖v‖H1(Ω), and hence it is a Hilbert space.
From this point of view, we are now able to formulate problem (P2) in order to precisely study it.

From this fact, we will need to the following hypothesis:

(H) :

{
f ∈ L2

(
0,T ; L2 (Ω)

)
(H.1)

ϕ ∈ H1 (Ω) (H.2)
.

3. Position of problem (P1)

In the rectangular area Q = Ω×(0,T ), and T < ∞, we consider the following linear problem (P2):


∂2u

∂t2
−a

∂2u

∂x2
= f (x,t) ∀(x,t) ∈ Q

u (x, 0) = ϕ (x) ∀x ∈ (0, l)
ut (x, 0) = ψ (x) ∀x ∈ (0, l)

∂u

∂x
(0,t) =

∫ l
0
k(x,t)g(ut)(x,t)dx ∀t ∈ (0,T )

∂u

∂x
(l, t) =

∫ l
0
k(x,t)g(ut)(x,t)dx ∀t ∈ (0,T )

(P2)

in which the hyperbolic equation is given as follows:

Lu =
∂2u

∂t2
−a

∂2u

∂x2
= f (x,t), (3.1)

with the initial conditions:

`u =

{
u(x, 0) = ϕ (x)

ut (x, 0) = ψ (x)
, x ∈ (0, l) ,

and with the integral condition of the second type:

∂u

∂x
(0,t) =

∫ l
0

k(x,t)g(ut)(x,t)dx, t ∈ (0,T ) ,

∂u

∂x
(l, t) =

∫ l
0

k(x,t)h(ut)(x,t)dx, t ∈ (0,T ) ,

where

k(x,t) ≥ 0 ∀(x,t) ∈ Q and 0 ≤ g(ut)(x,t) ≤ h(ut)(x,t) ∀(x,t) ∈ Q,

and

‖g(v)‖L2(Q) ≤ C‖v‖L∞(0,T,L2(Q)) ,



4 Int. J. Anal. Appl. (2023), 21:34

such that the space V = H1 (Ω) provided with the norm ‖v‖V = ‖v‖H1(Ω) is a Hilbert space. We are
now able to formulate the problem (P2), precisely to study it, according to the following hypothesis:

(H) :

{
f ∈ L2

(
0,T ; L2 (Ω)

)
(H.1)

ϕ ∈ H1 (Ω) (H.2)
.

Definition 3.1. The weak solution of problem (P2) is a function that satisfies:

• u ∈ L2
(

0,T ; H1 (Ω)
)
∩L∞

(
0,T ; H1 (Ω)

)
.

• u admits a strong derivative
∂u

∂t
∈ L2

(
0,T ; L2 (Ω)

)
.

• u (0) = ϕ,ut (0) = ψ.
• The following identity:

(utt,v) + a (ux,vx ) = (f ,v) + ux (l, t)v(l) −ux (0,t)v(0) ∀v ∈ V, ∀t ∈ [0,T ] .

3.1. Variational formulation. By multiplying the equation:

∂2u

∂t2
−a

∂2u

∂x2
= f (x,t) (3.2)

by an element v ∈ V , and the by integrating the result over Ω, we obtain:∫
Ω

∂2u

∂t2
·vdx −a

∫
Ω

∂2u

∂x2
·vdx =

∫
Ω

f ·vdx. (3.3)

By using the boundary conditions and using Green’s formula, ( 3.3) becomes

(utt,v) + a (ux,vx ) = (f ,vt) + ux (l, t)v(l) −ux (0,t)v(0), ∀v ∈ V, (3.4)

where (·, ·) denotes the scalar product L2 (Ω).

3.2. Study of the existence of weak solution of problem (P2). The demonstration of the existence

of the solution of problem (P2) can be discussed based on the Faedo-Galerkin method which consists

of carrying out the next three steps.

3.2.1. Step 1: Construction of the approximate solutions. As the space V is separable, then there

exists a sequence w1,w2, · · · ,wm having the following properties:

wi ∈ V, ∀i,
∀m,w1,w2, ...,wm are linearly independent,
Vm = 〈{w1,w2, ...,wm}〉 is dense in V.

(3.5)

In particular, we can say:

∀ϕ ∈ V =⇒ ∃(αkm)m ∈ IN∗, ϕm =
m∑
k=1

αkmwk −→ ϕ when m −→ +∞. (3.6)

∀ψ ∈ V =⇒ ∃(βkm)m ∈ IN∗, ψm =
m∑
k=1

βkmwk −→ ϕ when m −→ +∞. (3.7)



Int. J. Anal. Appl. (2023), 21:34 5

Now, the Faedo Galerkin’s approximation aims to search about a function in which

t 7→ um (x,t) =
m∑
i=1

gim (t) wi (x)

verifies {
um (t) ∈ Vm, ∀t ∈ [0,T ]

((um(t))tt ,wk) + A (um(t),wk) = (f (t),wk) ∀k = 1,m
, (P3)

for any integer m ≥ 1, where

((um(t))tt ,wk) =

((
m∑
i=1

gim (t) wi

)
tt

,wk

)

=

(
m∑
i=1

∂2gim
∂t

(t) wi (x) ,wk

)
=

m∑
i=1

(wi,wk)
∂2gim
∂t2

(t) ,

(3.8)

and

A (um(t),wk) = A

(
m∑
i=1

gim (t) wi,wk

)

= a

m∑
i=1

gim (t)


∫

Ω

∂wi
∂x

∂wk
∂x

dx −
∂wi
∂x

(l) wk(l) +
∂wi
∂x

(0) wk(0)




= a

m∑
i=1

gim (t)

∫
Ω

∂wi (x)

∂x

∂wk(x)

∂x
dx −a

m∑
i=1

gim (t)
∂wi
∂x

(l) wk(l)

+ a

m∑
i=1

gim (t)
∂wi
∂x

(0) wk(0) =

m∑
i=1

A (wi,wk) gim (t) .

(3.9)

In addition, we have

um(0) =

m∑
i=1

gim (0) wi (x) = ϕm =

m∑
i=1

αimwi (x).

and

u′m(0) =

m∑
i=1

g′im (0) wi (x) = βm =

m∑
i=1

βimwi (x).

We obtain consequently a system of first-order nonlinear differential equations:


m∑
i=1

(wi,wk)
∂2gim
∂t2

(t) + a
m∑
i=1

(
∂wi
∂x

,
∂wk
∂x

)
gim (t)

= (f (t),wk) + a
m∑
i=1

gim (t)
∂wi
∂x

(l) wk(l) −a
m∑
i=1

gim (t)
∂wi
∂x

(0) wk(0)

gim (0) = αim ∀i = 1,m.
g′im (0) = βim ∀i = 1,m.

, (P4)

From this view, we consider the vector:

gm = (g1m(t), · · · ,gmm(t)) , fm = ((f ,w1) , · · · , (f ,wm)) ,



6 Int. J. Anal. Appl. (2023), 21:34

coupled with the matrices:

Bm =
((
wi,wj

))
1≤i≤m
1≤j≤m

,Am =

((
∂wi
∂x

,
∂wj
∂x

))
1≤i≤m
1≤j≤m

,

and

Cm =

(
∂wi
∂x

(l) ·wj(l)
)

1≤i≤m
1≤j≤m

,Dm =

(
∂wi
∂x

(0) ·wj(0)
)

1≤i≤m
1≤j≤m

.

Now, we can immediately write problem (P4) in the matrix form as:

Bm

∂gm
∂t

(t) + aAmgm + aDmgm = fm + aCmgm

gm (0) = (αim)1≤i≤m

g′m (0) = (βim)1≤i≤m

,

As the matrix entries Bm are linearly independent (because it is a diagonal matrix), then det Bm 6= 0.
So, it is invertible, and then gm is the solution of the following states:


∂2gm
∂t2

(t) +
(
aB−1m Am + bB

−1
m Dm −aB−1m Cm

)
gm = B

−1
m fm

gm (0) = (αim)1≤i≤m .

g′m (0) = (βim)1≤i≤m .

(P5)

Now, it is easy to verify that this ordinary differential system has a solution where the matrix:

(
aB−1m Am + bB

−1
m Bm −aB

−1
m Cm

)
is of constant coefficients and the vector B−1m fm are continuous functions and majorized by integrable

functions on (0,T ). Consequently, we can conclude that there exists a tm that depends only on |αim|
and |βim|.

3.2.2. Step 2: A priori estimate. Herein, we intend to begin this step with state and prove the next

result.

Lemma 3.1. For all m ∈N∗, if
1

8a
> k2,

the solution um ∈ L2 (0,T ; Vm) of problem (P2) satisfies:

‖um‖L2(0,T ; H1(Ω)) ≤ c1,∥∥∥∥∂um∂t
∥∥∥∥
L2(0,T ; L2(Ω))

≤ c2,

where c1 and c2 are two positive constants independent of m.



Int. J. Anal. Appl. (2023), 21:34 7

Proof. By multiplying the equation of (P3) by gkm(t), and then by summing the result over k, we

obtain:
m∑
k=1

((um(t))tt ,wk) g
′
km(t) + a

m∑
k=1

(
∂um
∂x

(t),
∂wk
∂x∂t

)
g′km(t)

=

m∑
k=1

(f (t),wk) ·g′km(t) + a
m∑
i=1

gim (t)
∂wi
∂x

(l)

m∑
k=1

g′km(t)wk(l)

−a
m∑
i=1

gim (t)
∂wi
∂x

(0)

m∑
k=1

g′km(t)wk(0).

So, we obtain

((um(t))tt , (um(t))t) + a

(
∂um
∂x

(t),
∂um
∂x∂t

(t)

)
= (f (t), (um(t))t) + a

m∑
i=1

gim (t)
∂wi
∂x

(l)

m∑
k=1

g′km(t)wk(l)

−a
m∑
i=1

gim (t)
∂wi
∂x

(0)

m∑
k=1

g′km(t)wk(0).

Thus, we get

((um(t))tt , (um(t))t) + a

∥∥∥∥∂um∂x
∥∥∥∥2
L2(Ω)

= (f (t),um(t))

+ a

m∑
i=1

gim (t)
∂wi
∂x

(l)

m∑
k=1

g′km(t)wk(l) −a
m∑
i=1

gim (t)
∂wi
∂x

(0)

m∑
k=1

g′km(t)wk(0).

Integrating the above equality over 0 to t coupled wit using the Cauchy inequality with ε, i.e.

|ab| ≤
a2

2ε
+
εb2

2
,

we get

1

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Ω)

+
a

2
‖∇um‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+ a

m∑
i=1

gim (t)
∂wi
∂x

(l)

m∑
k=1

g′km(t)wk(l)

−a
m∑
i=1

gim (t)
∂wi
∂x

(0)

m∑
k=1

g′km(t)wk(0) +
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω) ,

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+ a
∂um
∂x

(l, t)
∂um
∂t

(l, t) −a
∂um
∂x

(0,t)
∂um
∂t

(0,t),

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

+ a

∫ τ
0

[(∫
Ω

k(x,t)g ((ut)m) (x,t)dx

)
∂um
∂t

(l, t)



8 Int. J. Anal. Appl. (2023), 21:34

−
(∫

Ω

k(x,t)h ((ut)m) (x,t)dx

)
∂um
∂t

(0,t)

]

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

+ a

∫ τ
0

[(∫
Ω

k(x,t)g ((ut)m) (x,t)dx

)
∂um
∂t

(l, t)

−
(∫

Ω

k(x,t)h ((ut)m) (x,t)dx

)
∂um
∂t

(0,t)

]
.

This means
1

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Ω)

+
a

2
‖∇um‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

+ a

∫ τ
0

[(∫
Ω

k(x,t)g ((ut)m) (x,t)dx

)
∂um
∂t

(l, t)

−
(∫

Ω

k(x,t)g ((ut)m) (x,t)dx

)
∂um
∂t

(0,t)

]

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

+ aCk

∥∥∥∥∂um∂t
∥∥∥∥
L∞(0,T ;L2(Ω))

[∫ τ
0

∫
Ω

∂2um
∂t∂x

dxdt

]

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

+ akC

∥∥∥∥∂um∂t
∥∥∥∥
L∞(0,T ;L2(Ω))

[∫
Ω

∂um
∂x

dx −
∫

Ω

∂ψm
∂x

dx

]
,

which yields

1

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Ω)

+
a

2
‖∇um‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

+
[(aCk)2

2δ

∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+ δ
[
‖∇um‖2L∞(0,T ;L2(Ω)) + ‖∇ϕm‖

2
L∞(0,T ;L2(Ω))

]]
+

1

2
‖ϕm‖2L2(Ω) +

a

2
‖∇ψm‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

ε

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Q)

(aCk)
2

2δ

∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+ δ‖∇um‖2L∞(0,T ;L2(Ω)) +
a

2
‖∇ψm‖2L2(Ω) + max(

1

2
,δ)‖ϕm‖2H1(Ω) ,



Int. J. Anal. Appl. (2023), 21:34 9

or
1

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Ω)

+
a

2
‖∇um‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

(
εCT

2
+

(aCk)
2

2δ

)∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+ δ‖∇um‖2L∞(0,T ;L2(Ω)) +
a

2
‖∇ψm‖2L2(Ω) + max(

1

2
,aCkδ)‖ϕm‖2H1(Ω) ,

(3.10)

where K = max
∫
Q
k2 (x,t) dxdt. Consequently, we obtain

1

2

∥∥∥∥∂um∂t
∥∥∥∥2
L2(Ω)

+
a

2
‖∇um‖2L2(Ω)

≤
1

2ε
‖f‖2L2(Q) +

(
εCT

2
+

(aCk)
2

2δ

)∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+ δ‖∇um‖2L∞(0,T ;L2(Ω)) +
a

2
‖∇ψm‖2L2(Ω)

+ max(
1

2
,δ)‖ϕm‖2H1(Ω) ,

which gives:(
1

2
−

(
εCT

2
+

(aCk)
2

2δ

))∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+
(a

2
−δ
)
‖∇um‖2L∞(0,T ;L2(Ω))

≤
1

2ε
‖f‖2L2(Q) +

a

2
‖∇ψm‖2L2(Ω) + max(

1

2
,δ)‖ϕm‖2H1(Ω)

By putting ε = 1
4CT

and δ = 4 (aCk)2, we get∥∥∥∥∂um∂t
∥∥∥∥2
L∞(0,T ;L2(Ω))

+ ‖∇um‖2L∞(0,T ;L2(Ω))

≤ C1
[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]
,

(3.11)

or

C1 =
max

(
2CT ,

a
2
, max( 1

2
, (aCk)

2
)
)

min

{
1

4
,
(
a
2
− 4 (aCk)2)

)} .
From (3.11), we can also get:∥∥∥∥∂um∂t

∥∥∥∥
L2(Ω)

≤
√
C1

[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]1
2
. (3.12)

By integrating (3.12) over [0,T ], we obtain:∥∥∥∥∥∥
τ∫
0

∂um
∂t

∥∥∥∥∥∥
L2(Ω)

≤
τ∫
0

∥∥∥∥∂um∂t
∥∥∥∥
L2(Ω)

≤ T
√
C1

[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]1
2
,



10 Int. J. Anal. Appl. (2023), 21:34

or ∥∥∥∥∥∥
τ∫
0

∂um
∂t

∥∥∥∥∥∥
L2(Ω)

≤ T
√
C1

[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]1
2
,

which implies:

‖um −ϕm‖L2(Ω) ≤ T
√
C1

[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]1
2
,

i.e.,
‖um‖2L2(Ω) + ‖ϕm‖

2
L2(Ω)

≤ TC1
[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]
+ 2‖um‖L2(Ω) ‖ϕm‖L2(Ω) .

By applying Cauchy inequality with γ, we get:

‖um‖2L2(Ω) + ‖ϕm‖
2
L2(Ω)

≤ TC1
[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]
+

1

γ
‖um‖2L2(Ω) + γ‖ϕm‖

2
L2(Ω) .

By putting γ = 2, we can have:

‖um‖2L2(Ω) ≤ 4TC1
[
‖f‖2L2(Q) + ‖∇ψm‖

2
L2(Ω) + ‖ϕm‖

2
H1(Ω)

]
. (3.13)

Now, it follows from (3.11) and (3.13) that the solution of the initial value problem for system (P4)

can be extended to [0,T ]. This confirms what we have demonstrated in the first step. Consequently,

when m → +∞ in (3.13), we obtain:{
um uniformly bounded in L2(0,T ; H1 (Ω))

(um)t uniformly bounded in L
2
(

0,T ; L2 (Ω)
) . (3.14)

�

3.2.3. Step 3: Convergence and the result of existence.

Theorem 3.1. There is a function u ∈ L2(0,T ; H1 (Ω)) ∩ L∞(0,T ; L2 (Ω)) with
∂u

∂t
∈

L2
(

0,T;L2 (Ω)
)
and a subsequence denoted by

(
umk
)
k
⊆ (um)m such that

 umk ⇀ u in L
2(0,T;H1 (Ω))

∂umk
∂t

⇀
∂u

∂t
in L2

(
0,T;L2 (Ω)

) ,
as m −→ +∞.

Proof. From Lemma 1.2, we might deduce that there are subsequences denoted respectively by
(
umk
)
,(

∂umk
∂t

)
of (um) and (um)t such that

umk ⇀ u in L
2(0,T ; H1 (Ω)) , (3.15)

and
∂umk
∂t

⇀ w in L2
(

0,T ; L2 (Ω)
)
. (3.16)



Int. J. Anal. Appl. (2023), 21:34 11

We know that according to Relikh-Kondrachoff’s theorem the injection of H1 (Q) into L2 (Q) is

compact. In addition, like the results of Rellich’s theorem, any weakly convergent sequence in H1 (Q)

has a subsequence which converges strongly in L2 (Q) . So, we can assert:

umk −→ u in L
2(Q) . (3.17)

On the other hand, from Lemma 1.3, there is a subsequence of
(
umk
)
k
, which is still denoted by umk,

converges almost everywhere to u such that

umk −→ u almost everywhere Q . (3.18)

It is still essential to demonstrate that w = ∂u
∂t
. This actually suffices to prove:

u(t) = ϕ +

t∫
0

w(τ)dτ. (3.19)

To this aim, we note that as

umk ⇀ u inL
2(0,T;L2 (Ω)) ,

then the proof of (3.19) is equivalent to prove that

umk ⇀ ϕ + χ in L
2(0,T ;L2 (Ω)) ,

which means

lim
(
umk −ϕ−χ,v

)
L2(0,T ;L2(Ω)) = 0, ∀v ∈ L

2(0,T;L2 (Ω)),

as

χ (t) =

t∫
0

w(τ)dτ.

In fact, by using the equality

umk −ϕmk =
t∫

0

∂umk
∂τ

dτ, for all t ∈ [0,T ] ,

with the help of using umk ∈ L
2
(

0,T ; Vmk
)
and

(
umk
)
t
∈ L2

(
0,T ; Vmk

)
, we can get:(

umk −ϕ−
t∫

0

w(τ)dτ,v

)
L2(0,T ;L2(Ω))

=


umk −ϕmk −

t∫
0

w(τ)dτ,v



L2(0,T ;L2(Ω))

+
(
ϕmk −ϕ,v

)
L2(0,T ;L2(Ω))

=


 t∫

0

(
∂umk
∂τ
−w(τ)

)
dτ,v



L2(0,T ;L2(Ω))

+
(
ϕmk −ϕ,v

)
L2(0,T ;L2(Ω)) ,



12 Int. J. Anal. Appl. (2023), 21:34

for all t ∈ [0,T ]. By virtue of part (ii) of Lemma 1.6, it comes(
umk −ϕ−

t∫
0

w(τ)dτ,v

)
L2(0,T ;L2(Ω))

=

t∫
0

(
∂umk
∂τ
−w(τ),v

)
L2(0,T ;L2(Ω))

dτ +
(
ϕmk −ϕ,v

)
L2(0,T ;L2(Ω)) ,

for all t ∈ [0,T ]. On the one hand, we have

lim
k−→∞

t∫
0

(
∂umk
∂τ
−w(τ),v

)
L2(0,T ;L2(Ω))

dτ = 0, (3.20)

for t ∈ [0,T ]. Besides, we have:

lim
k−→∞

(ϕm −ϕ,v)L2(0,T ;L2(Ω)) = 0, (3.21)

which implies:

lim
k−→∞

(
umk −ϕ−χ,v

)
L2(0,T ; L2(Ω)) = 0, ∀v ∈ L

2
(

0,T ; L2 (Ω)
)
.

�

Theorem 3.2. The function u of the Theorem (3.1) is the weak solution to the problem (P2) in the

sense of the definition 3.1.

Proof. From Theorem (3.1), we have shown that the limit function u satisfies the first two conditions

of the definition 3.1. Now we will demonstrate (iii). According to the Theorem 3.1, we have:

umk (0) ⇀ u (0) in L
2(Ω) .

On the other hand, we have

umk (0) −→ ϕ in L
2(Ω) ,

which implies:

umk (0) ⇀ ϕ in L
2(Ω) .

From the uniqueness of the limit, we get u (0) = ϕ. By using the same previous steps, we demonstrate

ut (0) = ψ. It remains to demonstrate (iv). To this aim, we have:

(utt,v) + a (u,v) = (f ,v) ∀v ∈ V, and ∀t ∈ [0,T ] .

Integrating (P3) over (0,T ), we obtain:

t∫
0

((um(t))tt ,wk) dτ +

t∫
0

a (um(t),wk) dτ =

t∫
0

(f (t),wk) dτ, (3.22)



Int. J. Anal. Appl. (2023), 21:34 13

for all k = 1,m and for all t ∈ [0,T ]. Using (3.13) and that Vm dense in V and passing to the limit
in (3.22), we get:

T∫
0

(utt,wk) dτ +

T∫
0

a (u,wk) dτ =

T∫
0

(f ,wk) dτ, ∀t ∈ [0,T ] .

This immediately implies that (iv) is verified. �

Corollary 3.1. The uniqueness of the solution of problem (P2) comes straight through the estimate

(3.11).

4. Weak solution of the nonlinear problem

Initially, we present the considered solution’s concept. For this purpose, we let v = v(x,t) be any

function of V such that

V =
{
v ∈ C1 (Q) , vx (l, t) = vx (0,t) = 0, t ∈ [0,T ]

}
.

By multiplying

∂2u

∂t2
−a

∂2u

∂x2
= f (x,t,u,ux )

by v and integrating the result over Qτ, we obtain∫
Qτ

∂2y

∂t2
(x,t) ·

∂v

∂t
(x,t)dxdt −a

∫
Qτ

∆y(x,t) ·
∂v

∂t
(x,t)dxdt

=

∫
Qτ

G (x,t,y,yx ) ·
∂v

∂t
(x,t)dxdt.

Now, by using integration by parts and the conditions on y and v, we get∫
Qτ

∂2y

∂t2
(x,t) ·

∂v

∂t
(x,t)dxdt + a

∫
Qτ

∂y

∂x
(x,t) ·

∂2v

∂x∂t
(x,t)dxdt

=

∫
Qτ

G(x,t,y,yx ) ·
∂v

∂t
(x,t)dxdt.

(4.1)

It then results from (4.1) that

A (y,v) =

∫
Qτ

G(x,t,y,yx ) ·
∂v

∂t
(x,t)dxdt, (4.2)

or

A (y,v) =

∫
Qτ

∂2y

∂t2
(x,t) ·

∂v

∂t
(x,t)dxdt + a

∫
Qτ

∂y

∂x
(x,t) ·

∂2v

∂x∂t
(x,t)dxdt.



14 Int. J. Anal. Appl. (2023), 21:34

Thus, it is the time to build a recurring sequence starting with y(0) = 0. The sequence
(
y(n)

)
n∈N

is defined as follows: Given the element y(n−1), then for n = 1, 2, 3, · · · , we can solve the following
problem: 



∂2y(n)

∂t2
−a∆y(n) = G

(
x,t,y(n−1),y

(n−1)
x

)
y(n) (x, 0) = 0

y
(n)
t (x, 0) = 0

y
(n)
x (0,t) = 0

y
(n)
x (l, t) = 0

. (P4)

According to the last linear problem, we fix the n each time. Problem (P4) admits then a unique

solution y(n) (x,t), which can be given by the Fadeo-Galarkin method. In this regard, we assume

z(n) (x,t) = y(n+1) (x,t) −y(n)(x,t).

As a result, we have the following new problem:


∂2z(n)

∂t2
−a∆z(n) = p(n−1)(x,t)
z(n) (x.0) = 0

z
(n)
t (x.0) = 0

z
(n)
x (0,t) = 0

z
(n)
x (l, t) dx = 0

, (P5)

or

p(n−1)(x,t) = G
(
x,t,y(n),y

(n)
x

)
−G

(
x,t,y(n−1),y

(n−1)
x

)
.

Multiplying

∂2z(n)

∂t2
−a∆z(n) = p(n−1)(x,t)

by z(n) and then integrating the result over Qτ yield:∫
Qτ

∂2z(n)

∂t2
(x,t) ·

∂z(n)

∂t
(x,t)dxdt −a

∫
Qτ

∆z(n)(x,t) ·
∂z(n)

∂t
(x,t) dxdt

=

∫
Qτ

p(n−1)(x,t) ·
∂z(n)

∂t
(x,t) dxdt.

If we apply an integration by parts for each term of the above equality, keeping in view the initial and

boundary conditions, we get:

1

2

∫
Ω

(
∂z(n)

∂t
(x,τ))2dx +

a

2

∫
Ω

(
∂z(n)

∂x
(x,t)

)2
dxdt =

∫
Qτ

p(n−1)(x,t) ·
∂z(n)

∂t
(x,t) dxdt.



Int. J. Anal. Appl. (2023), 21:34 15

When the Cauchy Schwarz inequality is applied to the second portion of the above equation, the

following result is obtained:

1

2

∫
Ω

(
∂z(n)

∂t
(x,τ))2dx +

a

2

∫
Ω

(
∂z(n)

∂x
(x,t)

)2
dxdt

6
1

2ε

∫
Qτ
| p(n−1)(x,t) |2 dxdt +

ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt,

or

1

2

∫
Ω

(
∂z(n)

∂t
(x,τ))2dx +

a

2

∫
Ω

(
∂z(n)

∂x
(x,t)

)2
dx 6

1

2ε

∫
Qτ
| G
(
x,t,y(n),y

(n)
x

)

−G
(
x,t,y(n−1),y

(n−1)
x

)
|2 dxdt +

ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt.

We deduce consequently that:

1

2

∫
Ω

(
∂z(n)

∂t
(x,τ))2dx +

a

2

∫
Ω

(
∂z(n)

∂x
(x,t)

)2
dx

6
k2

2ε

∫
Qτ

(| y(n) −y(n−1) | + | y(n)x −y
(n−1)
x |)2dxdt +

ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt

6
k2

2ε

∫
Qτ

(| z(n−1) | + | z(n−1)x |)2dxdt +
ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt

6
k2

ε

∫
Qτ

(| z(n−1) |2 + | z(n−1)x |2)dxdt +
ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt

6
k2

ε

∥∥∥z(n−1)∥∥∥
L2(0,T,H1(0,l))

+
ε

2

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt.

By multiplying by 2 and applying Grenwell’s Lemma, we get

∫
Ω

(
∂z(n)

∂t
(x,τ))2dx + a

∫
Ω

(
∂z(n)

∂x
(x,t)

)2
dx

6
k2

ε

∥∥∥z(n−1)∥∥∥
L2(0,T,H1(0,l))

+ ε

∫
Qτ

(
∂z(n)

∂t
(x,t)

)2
dxdt

≤
k2

ε
exp(εT )

∥∥∥z(n−1)∥∥∥
L2(0,T,H1(0,l))

.



16 Int. J. Anal. Appl. (2023), 21:34

Integrating over t yields:∫
QT

(
∂z(n)

∂t
(x,τ))2dxdt + a

∫
QT

(
∂z(n)

∂x
(x,t)

)2
dxdt

6
Tk2

ε

∥∥∥z(n−1)∥∥∥
L2(0,T,H1(0,l))

exp(εT ),

or ∫
QT

(
∂z(n)

∂t
(x,τ))2dxdt +

∫
QT

(
∂z(n)

∂x
(x,t)

)2
dxdt

6
Tk2 exp(εT )

ε min(1,a)

∥∥∥z(n−1)∥∥∥
L2(0,T,H1(0,1))

.

By putting

c =
Tk2 exp(εT )

ε min(1,a)
,

we get:

‖
∂z(n)

∂t
‖2L2(0,T,H1(0,1)) + ‖

∂z(n)

∂x
‖2L2(0,T,H1(Ω))6 c

∥∥∥z(n−1)∥∥∥2
L2(0,T,H1(Ω))

.

Thus, by applying Pointcarre, we have

‖ z(n) ‖2L2(0,T,H1(Ω))6 Tc
∥∥∥z(n−1)∥∥∥2

L2(0,T,H1(Ω))
,

and
n−1∑
i=1

z(i) = y(n).

According to the convergence criterion of the series
∞∑
n=1

z(n) that converges if |c| < 1, we obtain:

∣∣∣∣∣(Tk)
2

exp(εT )

ε min(1,a)

∣∣∣∣∣ < 1 ⇒ Tk
√

exp(εT )

ε min(1,a)
< 1.

Consequently, we get:

k <

√
ε min(1,a) exp(−εT )

T
.

Then (y(n))n converges to an element of L2(0,T,H1(Ω)), say y. Now, we will show that

lim
n−→∞

y(n)(x,t) = y(x,t)

is a solution to the problem (P5) by showing that y satisfies:

A (y,v) =

∫
Qτ

G(x,t,y,yx ) ·v(x,t)dxdt.

We therefore consider the weak formulation of the problem (P1) as follows:

A
(
y(n),v

)
=

∫
Qτ

∂2y(n)

∂t2
(x,t) ·

∂v

∂t
(x,t)dxdt +

a

2

∫
Qτ

∂y(n)

∂x
(x,t) ·

∂2v

∂t∂x
(x,t)dxdt.



Int. J. Anal. Appl. (2023), 21:34 17

From the linearity of A, we can have:

A
(
y(n),v

)
= A

(
y(n) −y,v

)
+ A (y,v)

=

∫
Qτ

∂2(y(n) −y)
∂t2

(x,t) ·
∂v

∂t
(x,t)dxdt +

a

2

∫
Qτ

∂(y(n) −y)
∂x

(x,t) ·
∂2v

∂t∂x
(x,t)dxdt

−
∫
Qτ

∂2y

∂t2
(x,t) ·

∂v

∂t
(x,t)dxdt +

a

2

∫
Qτ

∂y

∂x
(x,t) ·

∂2v

∂t∂x
(x,t)dxdt,

which implies

A
(
y(n) −y,v

)
=

∫
Qτ

∂2(y(n) −y)
∂t2

(x,t) ·
∂v

∂t
(x,t)dxdt

+
a

2

∫
Qτ

∂(y(n) −y)
∂x

(x,t) ·
∂2v

∂t∂x
(x,t)dxdt.

Now, by applying the Cauchy Schwartz inequality, the following results can be obtained:

A
(
y(n) −y,v

)
6‖ vt ‖L2(Qτ )

∥∥∥(y(n) −y)tt∥∥∥
L2(0,T,L2(Ω))

+
a

2
‖ vxt ‖L2(Qτ )

∥∥∥(y(n) −y)x∥∥∥
L2(0,T,L2(Ω))

.

Then, we can find

A
(
y(n) −y,v

)
6 C

(∥∥∥(y(n) −y)tt∥∥∥
L2(0,T,L2(Ω))

+
∥∥∥(y(n) −y)x∥∥∥

L2(0,T,L2(Ω))

)
×
(
‖ vt ‖L2(Qτ ) + ‖ vxt ‖L2(Qτ )

)
,

or

C = max
(

1,
a

2

)
.

Now, as y(n) −→ y in L2
(

0,T,H1 (0, l)
)
u H1 (Q), we get:

y(n) −→ y in L2 (Q) ,

y
(n)
t −→ yt in L

2 (Q) ,

y
(n)
x −→ yx in L2 (Q) .

Consequently, we note as n −→ +∞, we find:

lim
n−→+∞

A
(
y(n) −y,v

)
= 0.

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi-

cation of this paper.



18 Int. J. Anal. Appl. (2023), 21:34

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	1. Introduction
	2. The statement of the main problem
	3. Position of problem (P1)
	3.1. Variational formulation 
	3.2. Study of the existence of weak solution of problem (P2)

	4. Weak solution of the nonlinear problem
	References